# R is a commutative ring with unity and prime characteristic p, show that $\phi: R \to R\,\,/\,\, \phi(a) = a^p$ is a homomorphism

It's pretty obvious that $\phi(0) = 0$ and $\phi(1) = 1$ so those are all set.

Now I want to show that $\phi(a+b) = \phi(a) + \phi(b)$

or

$(a+b)^p = a^p + b^p$

for all $a,b \in R$.

however it's not immediate how to do this. $R$ being commutative doesn't help here at all, and the characteristic $p$ applies to addition not exponentiation..

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R being commutative doesn't help here at all

It does.

the characteristic p applies to addition not exponentiation

It does apply.

Use the binomial theorem to expand $(a + b)^{p}$ and look at the coefficients. Except for those of $a^{p}$ and $b^{p}$, they are multiples of $p$, hence reduces to zero under the assumption that $\mathrm{char} R = p$.

You see that the commutativity together with the finite characteristic of $R$ are crucial here.

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it's not immediately clear to me that the binomial coefficient are all multiples of p. –  terrible at math Mar 5 '14 at 6:02
Alright. $(a + b)^{p} = a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb \binom{p}{p - 1} + b^{p}$. Suppose $1 \leq k \leq p - 1$. Then we have $\binom{p}{k} = \frac{p (p -1)!}{k!(p - k)!} = p\times \frac{(p - 1)!}{k!(p - k)!}$. Note that there are no factors of $p$ appearing in $(p - 1)!, k!$ or $(p - k)!$. As this number must be an integer, it is divisible by $p$. Thus $\binom{p}{k} \equiv 0 \pmod p$. –  eltonjohn Mar 5 '14 at 6:20
Oh wow, I didn't think to pull that p out but that makes a lot of sense, thank you! Now we're just left with the first and last term.. perfect. I will try the multiplicative one on my own now! –  terrible at math Mar 5 '14 at 6:41
You are welcome. By the way a typo: $a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb \binom{p}{p - 1} + b^{p}$ should read $a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb + \binom{p}{p - 1}a b^{p -1} + b^{p}$ –  eltonjohn Mar 5 '14 at 6:47
Well this part was a lot easier. $\phi(ab) = (ab)^p$. so $(ab)(ab)(ab)(ab) . . . (ab)$. But R is commutative so this is the same as $(a)(a)(a)(a) . . . (b)(b)(b)(b)$ $p$ times each, or $a^p b^p$. QED question: when expanding arbitrary terms like that, is there a "correct" way to write that I have done this multiplication $p$ times? –  terrible at math Mar 5 '14 at 7:05

Use the binomial theorem to expand $(a+b)^p$ and then show that the binomial coefficients ${p \choose k}$ are divisible by $p$ for $0<k<p$.

Thanks Pedro for catching my careless error. For some reason, I'm having trouble getting to post comments right now, so I'm incorporating mine in my answer.

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$k=1$ too. ${}{}{}$ –  Pedro Tamaroff Mar 5 '14 at 4:14
it's not obvious to me that the binomial coefficient are divisible by p here –  terrible at math Mar 5 '14 at 6:05
Yes it is. Write down the factorial definition of the choose function. –  Joshua Biderman Mar 5 '14 at 6:15

Hint: Use the binomial theorem to expand $(a+b)^p$, and then use the fact that $R$ has characteristic $p$.

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i've expanded it but I don't see how I can use characteristic p. –  terrible at math Mar 5 '14 at 6:01